Multiplying Polynomials

In this tutorial, you’ll learn all about multiplying polynomials. But first, a quick look at a few basic, yet important concepts.

Polynomials - A Quick Introduction

Here’s a polynomial (with four terms).

The parts of the polynomial separated by the addition/subtraction symbols are called terms. And each term can be seen as having two parts, the coefficient and the variables (along with their exponents).

Anatomy of an algebraic term

Also, a polynomial having only a single term is known as a monomial. For example 2x4\hspace{0.2em} 2x^4 \hspace{0.2em} and 8a2b5\hspace{0.2em} 8a^2b^5 \hspace{0.2em}.

And a polynomial with exactly two terms is known as a binomial. For example, 2x4+8a2b5\hspace{0.2em} 2x^4 + 8a^2b^5 \hspace{0.2em} and x+4\hspace{0.2em} x + 4 \hspace{0.2em}.

Why monomials and binomials matter will become clear in a moment.

Multiplying Polynomials

Multiplying polynomials revolves around two things — multiplying monomials and applying the distributive property.

Once you are comfortable with these two concepts, multiplying polynomials becomes really simple. So, let’s dive in.

Multiplying Monomials

When multiplying two monomials, we multiply the coefficients together and identical variables together. And when multiplying identical variables, we add the exponents.

Let me explain using the following examples.

Example

Multiply the monomials.

(a)5x2y,−3xy3z(b)9pq,−pr2\begin{align*} &(a) \hspace{0.8em} 5x^2y \hspace{0.15em}, \hspace{0.5em} -3xy^3z \\[1em] &(b) \hspace{1em} 9pq \hspace{0.15em}, \hspace{0.5em} -pr^2 \end{align*}

Solution (a)

Step 1. We'll start by grouping constants together and identical variables together.

5x2y⋅−3xy3z=5⋅−3⋅x2⋅x⋅y⋅y3⋅z\begin{align*} & {\color{Red} 5} {\color{Teal} x^2} {\color{Orchid} y} \cdot {\color{Red} -3} {\color{Teal} x} {\color{Orchid} y^3} z \\[1.3em] = \hspace{0.4em} & {\color{Red} 5} \cdot {\color{Red} -3} \cdot {\color{Teal} x^2} \cdot {\color{Teal} x} \cdot {\color{Orchid} y} \cdot {\color{Orchid} y^3} \cdot z \end{align*}

Step 2. Now, we just need to multiply the entities in each group. And as I mentioned earlier, to multiply identical variables, we add the exponents.

Also, remember, if a variable doesn’t have a visible exponent, the exponent is 1\hspace{0.2em} 1.

5⋅−3⋅x2⋅x⋅y⋅y3⋅z=−15⋅x(2+1)⋅y(1+3)⋅z=−15x3y4z\begin{align*} & {\color{Red} 5} \cdot {\color{Red} -3} \cdot {\color{Teal} x^2} \cdot {\color{Teal} x} \cdot {\color{Orchid} y} \cdot {\color{Orchid} y^3} \cdot z \\[1.3em] = \hspace{0.4em} & {\color{Red} -15} \cdot {\color{Teal} x^{(2 + 1)}} \cdot {\color{Orchid} y^{(1 + 3)}} \cdot z \\[1.3em] = \hspace{0.4em} & {\color{Red} -15} {\color{Teal} x^3} {\color{Orchid} y^4} z \end{align*}

And that's our answer.

Solution (b)

Nothing too different here.

9pq⋅−pr2 = −9p(1+1)qr2= −9p2qr2\begin{align*} 9 {\color{Red} p} {\color{Teal} q} \cdot - {\color{Red} p} {\color{Orchid} r^2} \, &= \, -9 {\color{Red} p^{(1+1)}} {\color{Teal} q} {\color{Orchid} r^2} \\[1em] &= \, -9 {\color{Red} p^2} {\color{Teal} q} {\color{Orchid} r^2} \end{align*}

Multiplying a Monomial and a Binomial

Okay now that we can multiply two monomials, it’s time to step up and multiply a monomial with a binomial. Let me show how.

Here we have a monomial and a binomial (inside parentheses) multiplied together.

Multiplying polynomials - distributive property

So when we open the parentheses, we distribute the monomial across the terms inside. In other words, we multiply the monomial with each term of the polynomial within the parentheses.

Simple enough, right? Let's work on a couple of examples to make sure everything's clear.

Example

Simplify.

(i)4a,ab+7b2(ii)−xy2,3x2−2xy−y2+1\begin{align*} &(i) \hspace{1em} 4a \hspace{0.15em}, \hspace{0.5em} ab + 7b^2 \\[1em] &(ii) \hspace{0.4em} -xy^2 \hspace{0.15em}, \hspace{0.5em} 3x^2 - 2xy - y^2 + 1 \end{align*}

Solution (ii)

Let's start by writing out the problem in a more mathematical fashion.

4a⋅(ab+7b2)4a \cdot (ab + 7b^2)

Now, it's important to understand the role of parentheses here.

They are crucial because they tell us that the monomial 4a\hspace{0.2em} 4a \hspace{0.2em} is being multiplied with the whole binomial (both the terms). Without the parentheses, it would mean 4a\hspace{0.2em} 4a \hspace{0.2em} is being multipled with ab\hspace{0.2em} ab \hspace{0.2em} only and not ab+7b2\hspace{0.2em} ab + 7b^2.

Okay, now we distribute the monomial across the two terms of the binomial and each of the two resulting pairs.

4a(ab+7b2) = 4a⋅ab+4a⋅7b2= 4a2b+28ab2\begin{align*} {\color{Red} 4a} (ab + 7b^2) \, &= \, {\color{Red} 4a} \cdot ab + {\color{Red} 4a} \cdot 7b^2 \\[1em] &= \, 4a^2b + 28ab^2 \end{align*}

Done!

Solution (iiii)

Pretty much the same story here. Except that this time we need to distribute the monomial −xy2\hspace{0.2em} {\color{Red} -xy^2} \hspace{0.2em} across the four terms inside the parentheses. So,

−xy2⋅(3x2−2xy−y2+1) = −xy2⋅3x2−−xy2⋅2xy−−xy2⋅y2+−xy2⋅1= −3x3y2+2x2y3+xy4−xy2\begin{align*} & {\color{Red} -xy^2} \cdot (3x^2 - 2xy - y^2 + 1) \, \\[1em] = \, & {\color{Red} -xy^2} \cdot 3x^2 - {\color{Red} -xy^2} \cdot 2xy - {\color{Red} -xy^2} \cdot y^2 + {\color{Red} -xy^2} \cdot 1 \\[1em] = \, &-3x^3y^2 + 2x^2y^3 + xy^4 - xy^2 \end{align*}
−xy2⋅(3x2−2xy−y2+1) = −xy2⋅3x2−−xy2⋅2xy−−xy2⋅y2+−xy2⋅1= −3x3y2+2x2y3+xy4−xy2\begin{align*} & {\color{Red} -xy^2} \cdot (3x^2 - 2xy - y^2 + 1) \, \\[1.5em] = \, & {\color{Red} -xy^2} \cdot 3x^2 - {\color{Red} -xy^2} \cdot 2xy \\[0.5em] & \hspace{1em}- {\color{Red} -xy^2} \cdot y^2 + {\color{Red} -xy^2} \cdot 1 \\[1.5em] = \, &-3x^3y^2 + 2x^2y^3 + xy^4 - xy^2 \end{align*}

Multiplying Two Binomials

Alright, now things are getting exciting.

When multiplying two polynomials, each term in one polynomial is to be multiplied with each term in the other polynomial.

Now there are a couple of methods that help you do this in a systematic way. They are especially useful when you are just getting started because otherwise it's easy to make silly mistakes.

FOIL Method

The FOIL method helps you multiply two binomials together by making it an easy to follow process.

FOIL is an acronym for First Outer Inner Last. It tells you to multiply the first terms of the two binomials, then the outer terms, then inner terms, and finally, the last terms.

To understand how the method works, we’ll use it to multiply the following binomials.

(x−2)(x+3)(x - 2)(x + 3)

Let’s begin by labeling the four pairs of terms that we need to multiply together.

Multiplying polynomials - FOIL method
First-Outer-Inner-Last

And next, we’ll multiply each pair of terms and simplify.

That’s it. We have successfully multiplied the two binomials.

(x−2)(x+3) = x2+x−6(x - 2)(x + 3) \, = \, x^2 + x - 6

Let's do a couple more examples.

Example

Simplify using the FOIL method.

(i)(a+2b)(b+4)(ii)(x+y)2\begin{align*} &(i) \hspace{1em} (a + 2b)(b + 4) \\[1em] &(ii) \hspace{1em} (x + y)^2 \end{align*}

Solution (ii)

Remember, we multiply the first terms together, then the outer terms, then the inner ones, and finally the last terms.

(a+2b)(b+4)=  a⋅b+a⋅4+2b⋅b+2b⋅4=  ab+4a+2b2+8b\begin{align*} (& {\color{Red} a} + {\color{Teal} 2b} )( {\color{Orchid} b} + {\color{DarkOrange} 4} ) \\[1em] = \, \, & {\color{Red} a} \cdot {\color{Orchid} b} + {\color{Red} a} \cdot {\color{DarkOrange} 4} + {\color{Teal} 2b} \cdot {\color{Orchid} b} + {\color{Teal} 2b} \cdot {\color{DarkOrange} 4} \\[1em] = \, \, & ab + 4a + 2b^2 + 8b \end{align*}

That's it.

Solution (iiii)

Same story. Just remember to combine the like terms in the end (the two terms in orange in our case here).

(x+y)2  =  (x+y)(x+y)=  x⋅x+x⋅y+y⋅x+y⋅y=  x2+xy+xy+y2=  x2+2xy+y2\begin{align*} (x + y)^2 \, \, &= \, \,( {\color{Red} x} + {\color{Teal} y} )( {\color{Orchid} x} + {\color{DarkOrange} y} ) \\[1em] &= \, \, {\color{Red} x} \cdot {\color{Orchid} x} + {\color{Red} x} \cdot {\color{DarkOrange} y} + {\color{Teal} y} \cdot {\color{Orchid} x} + {\color{Teal} y} \cdot {\color{DarkOrange} y} \\[1em] &= \, \, x^2 + {\color{Blue} xy} + {\color{Blue} xy} + y^2 \\[1em] &= \, \, x^2 + {\color{Blue} 2xy} + y^2 \end{align*}

Grid (or Box) Method

The Grid Method is another way to make sure you multiply each term of the first polynomial with each term of the second polynomial.

And unlike the FOIL method, which worked for only binomials, the Box method works for polynomials with any number of terms.

To see how it works, let’s multiply these.

(x+2a)(x−3a)(x + 2a)(x - 3a)

So here are the steps.

Step 1.  Draw a grid and list the terms of the first polynomial vertically and those of the second horizontally (outside the grid).

Multiplying Polynomials - Box method, step 1

Step 2.  Fill each cell of the grid with the product of the terms marking the relevant row and column.

Multiplying Polynomials - Box method, step 2

Step 3.  Put together every term inside the grid to form a polynomial. This is the product.

x⋅x+x⋅−3a+2a⋅x+2a⋅−3a=x2−3ax+2ax−6a2\begin{align*} & {\color{Red} x} \cdot {\color{Teal} x} + {\color{Red} x} \cdot {\color{DarkOrange} -3a} + {\color{Orchid} 2a} \cdot {\color{Teal} x} + {\color{Orchid} 2a} \cdot {\color{DarkOrange} -3a} \\[1em] = \hspace{0.5em} & \hspace{0.5em}x^2 \hspace{0.4em} - \hspace{0.78em} 3ax \hspace{0.78em} + \hspace{0.42em} 2ax \hspace{0.42em} - \hspace{0.8em} 6a^2 \end{align*}

Step 4.  Add/Subtract the like terms to simplify the product.

x2−3ax+2ax−6a2=  x2−ax−6a2\begin{align*} & x^2 {\color{Blue} \hspace{0.2em}-\hspace{0.25em} 3ax + 2ax} - 6a^2 \\[1em] = \, \, & x^2 {\color{Blue} \hspace{0.2em}-\hspace{0.25em} ax} - 6a^2 \end{align*}
Examples

Simplify

(i)(2x−3)(5x−1)(ii)(a−b)2\begin{align*} &(i) \hspace{1em} (2x - 3)(5x - 1) \\[1em] &(ii) \hspace{1em} (a - b)^2 \end{align*}

Solution (ii)

Let's prepare the grid. As shown above, we will place the terms of the binomials outside the grid and then populate the grid with the products of term pairs.

And now we bring together the products from inside the grid and simplify to get our answer.

10x2−15x−2x+3=  10x2−17x+3\begin{align*} &10x^2 - 15x - 2x + 3 \\[1em] = \, \, & 10x^2 - 17x + 3 \end{align*}

Solution (iiii)

Again, we prepare the grid first.

And then add the product of each pair of terms to get the answer.

a2−ab−ab+b2=  a2−2ab+b2\begin{align*} &a^2 - ab - ab + b^2 \\[1em] = \, \, & a^2 - 2ab + b^2 \end{align*}

Which to Use — FOIL or Grid Method?

Both Foil and Grid methods are great, especially when you have just started with multiplying polynomials.

However, you don’t want to be always dependent on them. FOIL method cannot be used for polynomials with more than two terms, and the Grid method doesn’t look very elegant.

I (and almost everyone I know) use a strategy I want to show you next.

Multiplying Longer Polynomials

Take the first term of the first polynomial and multiply it with the second polynomial (every term). Then take the second term of the first polynomial and multiply it with the second polynomial. And so on.

Let me show you what I mean.

Examples

Simplify.

(i)(a+b)(2a−3b−4c)(ii)(x−y)(x2+xy+y2)\begin{align*} &(i) \hspace{1em} (a + b)(2a - 3b - 4c) \\[1em] &(ii) \hspace{0.65em} (x - y)(x^2 + xy + y^2) \end{align*}

Solution (ii)

Alright, the first step is to separate the terms in the first polynomial, multiplying the second polynomial with each term of the first.

(a+b)⋅(2a−3b−4c)=a⋅(2a−3b−4c)+b⋅(2a−3b−4c)\begin{align*} &( {\color{Red} a} {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} )\cdot (2a - 3b - 4c) \\[1em] = \hspace{0.4em} & {\color{Red} a} \cdot (2a - 3b - 4c) \\[1em] &\hspace{1em} {\color{Teal} + \hspace{0.2em}b} \cdot (2a - 3b - 4c) \end{align*}
(a+b)⋅(2a−3b−4c)=a⋅(2a−3b−4c)+b⋅(2a−3b−4c)\begin{align*} &( {\color{Red} a} {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} )\cdot (2a - 3b - 4c) \\[1em] = \hspace{0.4em} & {\color{Red} a} \cdot (2a - 3b - 4c) {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot (2a - 3b - 4c) \end{align*}

Now, we have two instances of a monomial multipled with a polynomial. That's something we have done before, right?

a⋅2a+a⋅−3b+a⋅−4c+b⋅2a+b⋅−3b+b⋅−4c {\color{Red} a} \cdot 2a {\color{Red} \hspace{0.2em}+ \hspace{0.2em} a} \cdot -3b {\color{Red} \hspace{0.2em}+ \hspace{0.2em} a} \cdot -4c \\[1em] \hspace{2.5em} {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot 2a {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot -3b {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot -4c
a⋅2a+a⋅−3b+a⋅−4c+b⋅2a+b⋅−3b+b⋅−4c {\color{Red} a} \cdot 2a {\color{Red} \hspace{0.2em}+ \hspace{0.2em} a} \cdot -3b {\color{Red} \hspace{0.2em}+ \hspace{0.2em} a} \cdot -4c {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot 2a {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot -3b {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}b} \cdot -4c

Next, we multiply each pair. So,

2a2−3ab−4ac+2ab−3b2−4bc {\color{Red} 2a^2 - 3ab - 4ac} {\color{Teal} \hspace{0.2em} + \hspace{0.2em} 2ab - 3b^2 -4bc}

And finally, combine the like terms (−3ab-3ab \hspace{0.2em} and +2ab\hspace{0.2em} +2ab \hspace{0.2em}) to get the simplified answer.

2a2−ab−4ac−3b2−4bc2a^2 - ab - 4ac - 3b^2 - 4bc

That's our answer.

It may look a bit intimidating at first and you may need some practice to get the hang of it. But trust me, it get easier and is totally worth the effort.

Solution (iiii)

Alright, we'll be following the same steps we had for the last one. Let's go.

(x−y)⋅(x2+xy+y2)=x⋅(x2+xy+y2)−y⋅(x2+xy+y2)=x⋅x2+x⋅xy+x⋅y2−y⋅x2−y⋅xy−y⋅y2=x3+x2y+xy2−x2y−xy2−y3=x3−y3\begin{align*} &( {\color{Red} x} {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} )\cdot (x^2 + xy + y^2) \\[1.75em] = \hspace{0.4em} & {\color{Red} x} \cdot (x^2 + xy + y^2) \\[1em] &\hspace{1em} {\color{Teal} - \hspace{0.2em}y} \cdot (x^2 + xy + y^2) \\[1.75em] = \hspace{0.4em} & {\color{Red} x} \cdot x^2 {\color{Red} \hspace{0.2em}+ \hspace{0.2em} x} \cdot xy {\color{Red} \hspace{0.2em}+ \hspace{0.2em} x} \cdot y^2 \\[1em] &\hspace{1em} {\color{Teal} - \hspace{0.2em}y} \cdot x^2 {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} \cdot xy {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} \cdot y^2 \\[1.75em] = \hspace{0.4em} & {\color{Red} x^3 + x^2y + xy^2} \\[1em] &\hspace{1em} {\color{Teal} -x^2y - xy^2 - y^3} \\[1.75em] = \hspace{0.4em} &x^3 - y^3 \end{align*}
(x−y)⋅(x2+xy+y2)=x⋅(x2+xy+y2)−y⋅(x2+xy+y2)=x⋅x2+x⋅xy+x⋅y2−y⋅x2−y⋅xy−y⋅y2=x3+x2y+xy2−x2y−xy2−y3=x3−y3\begin{align*} &( {\color{Red} x} {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} )\cdot (x^2 + xy + y^2) \\[1em] = \hspace{0.4em} & {\color{Red} x} \cdot (x^2 + xy + y^2) {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} \cdot (x^2 + xy + y^2) \\[1em] = \hspace{0.4em} & {\color{Red} x} \cdot x^2 {\color{Red} \hspace{0.2em}+ \hspace{0.2em} x} \cdot xy {\color{Red} \hspace{0.2em}+ \hspace{0.2em} x} \cdot y^2 {\color{Teal} - \hspace{0.2em}y} \cdot x^2 {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} \cdot xy {\color{Teal} \hspace{0.2em}- \hspace{0.2em}y} \cdot y^2 \\[1em] = \hspace{0.4em} & {\color{Red} x^3 + x^2y + xy^2} {\color{Teal} -x^2y - xy^2 - y^3} \\[1em] = \hspace{0.4em} &x^3 - y^3 \end{align*}

And with that, we come to the end of this tutorial on multiplying polynomials. Until next time.